Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class P, class Q>
inline void smin(P &a, Q b) {
  if (b < a) a = b;
}
template <class P, class Q>
inline void smax(P &a, Q b) {
  if (a < b) a = b;
}
const int maxn = 100 + 10;
int n, m;
int g[maxn][maxn];
vector<pair<int, int> > ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0, i__n = (int)(n); i < i__n; ++i)
    for (int j = 0, j__n = (int)(m); j < j__n; ++j) cin >> g[i][j];
  int mn = 999;
  for (int i = 0, i__n = (int)(n); i < i__n; ++i)
    for (int j = 0, j__n = (int)(m); j < j__n; ++j) smin(mn, g[i][j]);
  if (n < m)
    for (int x = 0, x__n = (int)(mn); x < x__n; ++x)
      for (int i = 0, i__n = (int)(n); i < i__n; ++i)
        ans.push_back(pair<int, int>(0, i));
  else
    for (int x = 0, x__n = (int)(mn); x < x__n; ++x)
      for (int j = 0, j__n = (int)(m); j < j__n; ++j)
        ans.push_back(pair<int, int>(1, j));
  for (int i = 0, i__n = (int)(n); i < i__n; ++i)
    for (int j = 0, j__n = (int)(m); j < j__n; ++j) g[i][j] -= mn;
  for (int i = 0, i__n = (int)(n); i < i__n; ++i) {
    mn = 999;
    for (int j = 0, j__n = (int)(m); j < j__n; ++j) smin(mn, g[i][j]);
    for (int x = 0, x__n = (int)(mn); x < x__n; ++x)
      ans.push_back(pair<int, int>(0, i));
    for (int j = 0, j__n = (int)(m); j < j__n; ++j) g[i][j] -= mn;
  }
  for (int j = 0, j__n = (int)(m); j < j__n; ++j) {
    mn = 999;
    for (int i = 0, i__n = (int)(n); i < i__n; ++i) smin(mn, g[i][j]);
    for (int x = 0, x__n = (int)(mn); x < x__n; ++x)
      ans.push_back(pair<int, int>(1, j));
    for (int i = 0, i__n = (int)(n); i < i__n; ++i) g[i][j] -= mn;
  }
  bool ok = true;
  for (int i = 0, i__n = (int)(n); i < i__n; ++i)
    for (int j = 0, j__n = (int)(m); j < j__n; ++j)
      if (g[i][j]) ok = false;
  if (ok) {
    cout << (int((ans).size())) << endl;
    for (pair<int, int> p : ans)
      cout << (p.first ? "col" : "row") << ' ' << p.second + 1 << endl;
  } else
    cout << -1 << endl;
  return 0;
}